What is the integral of (9-y)sqrt(y) ?

The integral of (9-y)sqrt(y) is 6y^{3/2}-2/5 y^{5/2}+C

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integral of (9-y)sqrt(y)

\int (9 - y) y d y

6 y^{\frac{3}{2}} - \frac{2}{5} y^{\frac{5}{2}} + C

Solution steps

\int (9 - y) y d y

Expand

(9 - y) y : 9 y - y^{\frac{3}{2}}

= \int 9 y - y^{\frac{3}{2}} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 9 y d y - \int y^{\frac{3}{2}} d y

\int 9 y d y = 6 y^{\frac{3}{2}}

\int y^{\frac{3}{2}} d y = \frac{2}{5} y^{\frac{5}{2}}

= 6 y^{\frac{3}{2}} - \frac{2}{5} y^{\frac{5}{2}}

Add a constant to the solution

= 6 y^{\frac{3}{2}} - \frac{2}{5} y^{\frac{5}{2}} + C

n = 1 \sum \infty (cos (24))^{n}

x \to 0 lim (\frac{x ^{2}}{2} - \frac{2}{x})

f (x) = arccos (\frac{x}{2})

\frac{d}{d y} (x y + y)

(- 5, 5), (3, - 4)

What is the integral of (9-y)sqrt(y) ?
The integral of (9-y)sqrt(y) is 6y^{3/2}-2/5 y^{5/2}+C